To determine the absolute maximum value of the function $f(x) = x^4 + 8x^3 + 16x^2 + 8$ on the closed interval $[-5, -1]$, we need to follow these steps:

1. Find the derivative of $f(x)$: $f'(x) = 4x^3 + 24x^2 + 32x$

2. Set $f'(x) = 0$ to find critical points within the interval:

   We solve: $4x^3 + 24x^2 + 32x = 0$

   Factor out $4x$: $4x(x^2 + 6x + 8) = 0$ Then, further factor $x^2 + 6x + 8$: $4x(x + 4)(x + 2) = 0$

   This gives the solutions: $x = 0, \quad x = -4, \quad x = -2$

   Since we are only interested in values within the interval $[-5, -1]$, we discard $x = 0$ and keep $x = -4$ and $x = -2$.

3. Evaluate $f(x)$ at the endpoints and critical points in the interval:

   • At $x = -5$: $f(-5) = (-5)^4 + 8(-5)^3 + 16(-5)^2 + 8 = 625 - 1000 + 400 + 8 = 33$

   • At $x = -1$: $f(-1) = (-1)^4 + 8(-1)^3 + 16(-1)^2 + 8 = 1 - 8 + 16 + 8 = 17$

   • At $x = -4$: $f(-4) = (-4)^4 + 8(-4)^3 + 16(-4)^2 + 8 = 256 - 512 + 256 + 8 = 8$

   • At $x = -2$: $f(-2) = (-2)^4 + 8(-2)^3 + 16(-2)^2 + 8 = 16 - 64 + 64 + 8 = 24$

4. Determine the absolute maximum:

   By comparing these values, we see that the maximum value is: $\max\{33, 17, 8, 24\} = 33$

Conclusion:

The absolute maximum value of $f(x)$ on the interval $[-5, -1]$ is $\boxed{33}$, which occurs at $x = -5$.

Would you like details on any step, or do you have further questions?

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Here are some related questions you might find useful:

1. How do you determine if a critical point is a maximum or minimum?
2. How would you analyze this function on a different interval?
3. What is the difference between absolute and relative maxima?
4. How can you use the second derivative test to confirm a maximum?
5. What are other methods for solving polynomial equations like this one?

Tip: Always evaluate your function at both critical points and endpoints when looking for absolute extrema on a closed interval.